VII transversal

A .Genericityoftransversalityweknow what it means for a mapf : M -7N

to be transverse to a submanifold S ofN

and that it f is transverse to S (write f hts)then f

- '( s) is a submanifold of M of

codimension = Lookin S in N.

exercise: if an *or and flyn :3M→N is ht to S

and f : M → N is ht to S

then f- '

( s ) is a submanifold of M

with 2 f-

4S) C 2M

How easy is it to find ht maps ?

That:-let Mi N

,X be smooth manifolds

(M can have boundary)

let s c N be a submanifold.

if F : Mx X → N is transverse to S.ena÷:¥'seII"meawf-×

'. M -7 N : pt ftp.x) is IT to S

Proofs : we assume 2M =0 and leave

2M For as an exercise

note : I = F- 'l s ) is a submanifold of M xx

let IT :M XX → X : (pix) t X

claim :

x a regular value of Tls ⇒ f-×ht s

m item

I E. f :Xxt a

TY x

given the claim we are done by sard's Tha !

Pf of claim=

:

suppose x is a regular value of Theneed to see fx ht s

suppose Cp, x ) E -2 so ftp.xl-s ES

Fht s ⇒ Vv C- TSN ,3- we Tip.⇒ CMXX

)

St. V- d Ep.×, Cw) E Tss

we need to Aid u c- Tp M set.

V - dpcu) E T St- facp)

now Tcp,×)(M x X ) = Tp Mx Tx XU U

w = (ai b)

if b -- O,then consider

Mit Mx Xp t (p, x )

we know dip Ca) = Caio )

so ldfx)p la ) = deo i)p Ca)= d Fcp.⇒ (dip cat )= dEp.⇒ la . 01 = d Ep,×, la .b)

= deep,×, Iw)

and so V-Cdt)pca) E Tss

and we are done

now in general, we know thatdip

.⇒ ftp.xy-2) = Tx X

since x is a regular value

and d Typ.×,'

. Tcp.⇒ (M xx) → Tx Xl l

Tp M x TxX

is just projection

so F Cc , b) E Tcp,×, I set. d Typ.×,Kib) = b

note : since Cc,b) C- Tcp

.×,-2 we know

d Fcp.×, Cc . b) c- Tss

so V- d Ep.* ( a - c, o) -- V- deep,×, Kaist - cab))

=p - deep.

aid) + dE.*kid--

C- Tss E Tss

: . v- deep.⇒ la - c, o) E Tssand we are done by above L#

THAI:I

let M ,N be smooth manifolds (m possibly with 2)and s a submanifold ofN

for any smooth mapf :M →N there is aT.fi#sa::othi.:aonfsin-wW

Proof : we can think of Nc IRK for some k

let X be the unit ball in Rh

now set F :M xx → N

(Y . x)t q (fly)t x )wcall Ely

,x)

where q : NCN) → NT tubular nbhd ofN in Rh

note : Tannenxx) - Tim xtqyxnd-ETnyyynnkd-I.amclearly d F- is a submersion (just consider

vectors in Tx x)

and by Th'VI. 4 dg is a submersion

:. DF is a submersion !

so F is transverse to any submanifold of N

:. by Th' l F a dense set of x c- X s.t.

fact -- Fl ; x ) is It to S

now let G : Mx lo. D→N

④ t ) l→ Fly, tx) for xasf

then this is a homotopy from Gly, ol = Fly, o) = fly)

to Gly, x) =Fly, x) = fxcy) ht s ¥7

Thi :#

let MiN be smooth manifolds (M possibly with 2)S a smooth submanifold ofN

f : M →N a smooth map

ifI a closed set C ofM on which f is IT toS(and flan hts on C ), then÷÷÷÷÷÷÷¥:÷¥:

Idea of proof .-

step there is an open neighborhood Uof C set.

f- lard ft ) is ht to S on U

Ain't ht is an"

open"

condition

see proof of stability the

step. there is a function T :M→ IR Sf.

3- open sets ( c U'

cu"

c U and

Y= O on T'

and

8=1 outside U' '

Hint : Cor VI. 3

Steps : If F : Mx X→ Yr is the function from

last proof, then setG : M xx→N

Cy, x ) 1-7 F Cy, 8'

ly) x)

note : i) where 8=1 ( outside U"),

G ht s since F is

c) where V -- O ( inside U'),

G Cy,a = Fly, o) = fly) ht s

since f is on U'

3) exercise G is transverse to S

everywhere ( re. on U' '

- U)

now proof follows as proof of That2 ¥7

B. 1-Manifoldsandappl.cat

The--

:

every compact connectedl-manifold is

ditfeomorphcitoco.io#Remark-:we will not prove this

It is not too hard,see Guillemin & Pollack

Tha 4 .I

-

let M be a smooth compact manifold with boundarythere is no continuous retraction of M to 2M2Enomapf:M→2Ms¥f=idon2

Proof : suppose there is a retraction f : MMM

assume f is smooth (for now)F x c- 2M that is a regular value of f and flyµ

so f- '(x) = S is a submanifold of M and as c 2M

{A closed in 2M so s closed in M

M compact so S is compact'

.. S = union of intervals and S

' 's

note: f- 'Cx) n 2M = {x) since f- = id on 2M

so S has one boundary point !but compact t- manifolds must have an even

number of boundary points XO

so f does not exist !

now suppose f- is only continuous

by THEVI.3 F a nbhd of 2M in M

diffeomorphic to 2M x [o. E)

now setf : zme, go. c) → 3M x to

,E)

②it) Tf (x , g CH)

gCf),where

¥44

"

42 C

and extend F to rest ofM by using t

note: F is smooth on 2Mx Lo , Ely)

(just projection to 2n)

so by Th' -VI. 5 we can homotop I to

F rel 2M so that F is smooth

thus I is a smooth retraction *#

Cor5(Browerfixed-pointeh

any continuous map f : B"

→ B"

nasatieas.te.on.ee#nes::i:::mt

Proof : assume not

then fix) and x define a ray r starting at fix) and

going through x×

let gan - ingen:.int q¥gcx)

as shown in picture

claim :g Cx) is continuous

it true then this * THE 4 since gcxtx for x EJB"

to see the claim is true note rix is parameterized

by f-f) text t tx tZo

the ray intersects 213"

at

C-t)'

II fix)Ift ZHI-H lxofcxl) t t' 11×112=1

SO

(11×112-2 x.Hx) t HABIT) t't 2 ( x. Hx) - Htmlf) tellHALF 1=0→

o

HX - fix) H'

t't 2 (x. fix) - Kfc t t (HfCHIT- I) = O

--put = polynomial in t

so p has two roots

and we know p co)= Hf Cx) H

'- l E O

p Ll) = 11×42 E O

but put > 0 as t→ Ix

so one root E O and one 21

quadratic formula says the positive root Exdepends continuously on x

i. g Cx, = 4- tx) flat tax is continuous

.LT#C.Mod2inIerseclrySuppose M .

N are manifolds5 a submanifold of N

dim M t dim S = din N

M,s have no boundaryseed in N

M compact

Then givea f : M →N ,

we can use The 2 to

homotop f to f , so that f,hts

note: fi'

( s ) = f,(M ) n s = { finite set of points )

to- devil submanifold

compact since M is and

S is closed

wits

Example:

I H

im f s im f s

00 00

The intersections is

Idf , s ) -- # f?(s ) mod 2

Three :-

Iz ft , s ) is well -definedi.eitfih.ca?sfIItII.sProof : let F : M x [o

,I ] → N be the homotopy

from f,to fz

by Tha 3 we may find a homotopy F st. Ent s

now F- '( s ) = compact t- manifold

= I s'u V

,u

. . .

u rnw

n-

M x (o,i )

arcs with end

points on Mxfo, B

note f,- '

(s ) v fi'

( s) = 2 (F- '

Cs ) ) ← even # of pts

so # f,

- '

(s) = # fi'

( s) mod 2↳#

Remarks .

-

i ) given any f, ,fz : M → N homo topic then

Idf, , s ) = Izlfz , s)

(since homotopy is an equivalence rel')

2) given S, , Sz submanifolds of N set

.

dinis, t dim S,= doin N

( we say S, , Sz have complementary doin)

suppose. S, , S, closed and compact

then Iz( s

, ,Sa) is defined to be Ili, Sa)

where i :S,→ N is inclusion

Th 17 :

-

suppose M ,N,S and f : M→ N as above

it 3 a compact manifold W set.

2W-- M andt÷÷:;::oeProof : homotop F to E that is transverse to S

note : Iz (f. s) = I ( F- law , s ) since f and Flaware homotopic

nowI - '( s) is a compact t -manifold

so 2 F- 'l s) = even # of points

11

⇐bj'

Is )

so Iz ( Flow , 5) = O EA

example: consider T'= s

'x s

'

let s = S'

x Ipt }

and f : s'→ T2

0 1-7 (pt, o)

① to

note f ht s and f- '(s ) = {pt }

so Idf , s ) = I

thus f does not extend over D'

( or any compact surface )

Intuitively clear but not easy to show !

exercise: it show 52 is not diffeomorphic to T'

by showing any f : s'→ 5 can be

extended to DZ

2) Show IRP'

is not diffeomorphic to5

Hint : consider"self - intersection "of

s'a 5→ IRPZequator. g

suppose now M ,N are two manifolds of the same

dimension and without boundary

assume M is compact and

N is connected

given any map f:M →N we call Idf, Ep))

for any p EN the degd2 of f

and denote it deg.it)

Thad :

-

F~%7e.thgeg.FI??Ien7dem?eaFdHpih=IiHp#for the proof we need a few observations

first since deg, ft) is defined in terms of Idf . 1ps) and

Idf, {p)) is unchanged if we homotop f we see

GI :-

homotopicmapshauethesamedeg.ee#we also need

lemma 10-

:

-

With notation as above,for any regular value p of f

F an open nbhd U of p st.

f-'

(u) = U,U. .

.u Uk%fEeothwe.ph?q?disiointandf:ui-su.sa#'

Proof of The 8 .

-

given p EN we can homo top f to be transverse

to p l so p is a regular value of f)

now we have U, , . . . , uh as in lemma 9

so for any q E U , f- '

(q ) has k points

i. the function

N → Z

pt Idf , Ip))

is locally constantsince N is connected it is thus constant

(exercise)

EH

Proof of Lemma 9#let {pi, . . ., ph }

- f - Yp) (we know f- '(p) is a compact o - manifold

so a finite number ofpoints)

each pi has a neighborhood Wi such thatf- Iw,

: Wi → f-(Wi)

is a diffeomorphism (by inverse function tha)

since M is Hausdorff we can assume Wi are disjoint

note X= f- (M -EW;) is compact (since M is)

set U =(f- (4) n . . . nflhk)) - Xneed to remove x since

some pts in M- U W; might

this is an open set and peu map to h Hwi)

now set Vi = Win f- '(u)

Hearty flu,

: u;→ U is a diffeomorphism

and UVi e f- ' (u)

now if XE f- '

(u) then fix) E U

so f-CH EX = f- (M-at,wi)

i. xe U wi ⇒ x E wi some i

and so X E wi nf-Yu) = Vi

thus U Vi = f-Yu)EH

The l l :-

H M - 2W,W compact, and f : M→N can be

extorted to a map F : W→ N thendegzCf€Proofs : take a regular value p of f

so the degz (f) = # (f-

Yp)) mod 2

we can homo top F rel 2W (re. don't change f)so that FAT {p }

now F-'

Ip ) is a compact t-manifold with boundaryf-

'

(p) so f- '

(p) is an even number of pts#=

Cool application :

Th"veryTmplexpolynomialotodddegreehasaroRemarks :

i ) similar proof with non -mod 2 degree provesgeneral result ( any positive degree)

2) The proof is a simple case of a very powerful idea

Thecontinvitymethod : if you can't solve an

equation, homotop it to one you can solve

and then argue original one solvable too !

Proofs : let plz ) be an odd degree polynomialit plz ) has no zeros then

P CZ)

ET: E → s

'

is well -defined

now consider

Pth )- t pH ) +4- t) zm

= t (Exam.,tm

- 't

. . .tao ) t ( I- t) Zm

( can assume p is monic)

note : away from origin pt has a zero

⇒

Pelz)Fm

does

but P¥# = It t ( am-1¥ +

. . .

t 9¥)-

as2-→ no

,

→ o

④

so it Dr is the disk of radius r about origin in G

then for large r , I⑦ I k 1 on 3Dr

so PfHH o on 2Dr

- :Pt CZ)Tptzt

"

27g,→ s

'

is well- defied

and gives a homotopy from Yp¥I¥o ¥7,↳,

you can easily check deg, ( FT ) = m I 0 mod 2it m odd

so degt ff¥↳! * oand PCI

pay ↳ can 't be extended over Dr

so from above plz) must have a root in Dr#